Answer:
Zn°(s) + Fe⁺²(aq) => Zn⁺²(aq) + Fe°(s)
Explanation:
Molecular Equation:
Zn°(s) + Fe(NO₃)₂(aq) => Zn(NO₃)₂(aq) + Fe°(s)
Ionic Equation:
Zn°(s) + Fe⁺²(aq) + 2NO₃⁻(aq) => Zn⁺²(aq) + 2NO₃⁻(aq) + Fe°(s)
Net Ionic Equation: => Drop NO₃⁻ as spectator ion
Zn°(s) + Fe⁺²(aq) => Zn⁺²(aq) + Fe°(s)
Net Primary Productivity ... the amount of biomass present in an ecosystem at a particular time .... Explain why a slow growing forest can have a very low NPP and yet store a massive amount of biomass.
Answer : The half-life at this temperature is, 3.28 s
Explanation :
To calculate the half-life for second order the expression will be:
![t_{1/2}=\frac{1}{k\times [A_o]}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5Ctimes%20%5BA_o%5D%7D)
When,
= half-life = ?
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial concentration = 0.45 M
k = rate constant = 
Now put all the given values in the above formula, we get:
Therefore, the half-life at this temperature is, 3.28 s
Answer:
The answer to your question is: Excess oxygen = 2.3 mol
Explanation:
Data
ZnS = 5 mol
O2 = 9.8 mol
Excess reactant = ?
Balanced reaction
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
MW ZnS = 65 + 32 = 97 x 2 = 194 g
MW O2 = 16 x 6 = 96 g
2 mol of ZnS ------------------- 3 mol O2
Ratio from the reaction = 3 mol O2/ 2 mol ZnS
= 1.5
Ratio from the quantities in the experiment = 9.8 mol O2 / 5 mol of ZnS
= 1.96
Excess reactant = Oxygen because the ratio increases
2 mol of ZnS ------------------- 3 mol O2
5 mol of ZnS ------------------- x
x = (5 x 3) / 2
x = 7.5 mol of O2
Excess Oxygen = 9.8 mol - 7.5 mol
Excess oxygen = 2.3 mol
From the equation, we see that the molar ratio of Fe : S required is:
8 : 1
The moles of Fe present are: 9.42/56 = 0.168
Moles of S = 68/(32 * 8) = 0.265
The molar ratio is:
1 : 1.6
Therefore, iron is the limiting reactant as it is present in a ratio lower than that required. The ratio of
Fe : FeS is
1 : 1
So 0.168 moles of FeS will form. The mass of FeS will be:
Mass = 0.168 * (56 + 32)
Mass = 14.78 grams
14.78 grams of FeS will be formed.
